\(\left|x+3\right|=\left|1-x\right|\)
TH1 : Với \(x+3\ge0\Leftrightarrow x\ge-3\)
\(\Leftrightarrow x+3=1-x\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-1\) ( Thỏa mãn )
TH2 : Với \(x+3< 0\Leftrightarrow x< -3\)
\(\Leftrightarrow-x-3=1-x\)
\(\Leftrightarrow0x=4\) ( Không thõa mãn )
Vậy \(S=\left\{-1\right\}\)
Chúc bạn học tốt
\(\dfrac{\left(x+1\right)\left(x+2\right)-\left[\left(x+1\right)-x\right]}{\left(x+2\right)\left[\left(x+1\right)^2-x\right]}-\dfrac{\left(x+1\right)+2-\left(x+1\right)\left[\left(x+1\right)^3+1\right]}{\left(x+1\right)^3+1}\)
Giải phương trình
a) \(2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)
b) \(\left(2x-2\right)^3=\left(x+1\right)^2+3\left(x-2\right)\left(x+5\right)\)
c) \(\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)
d) \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-2\right)-8\)
a) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
b) \(\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=6\)
c) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=840\)
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
Tìm x biết:
a) \(2x\left(5x-3\right)-\left(2x-1\right)\left(2x+1\right)-8x^2+6x\)
b) \(\left(x^2+2\right)\left(x-1\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
Các chế ơi, còn 10 câu nữa thui, sắp hết rùi.
FIGHTING!
JIAYOU!
HWAITING!
GAMBATTEYO!
Giải các phương trình sau
16) \(3\left(x+5\right)\left(x+6\right)\left(x+7\right)=8x\)
17) \(\left(x+6\right)^4+\left(x+4\right)^4=82\)
18) \(\left(x^2+6x+10\right)^2+\left(x+3\right)\left(3x^2+20x+36\right)=0\)
19) \(2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\left(x^3-1\right)\)
20) \(\left(x+2008\right)^4+\left(x+2009\right)^4=\dfrac{1}{8}\)
21) \(x^4+18x=13x^2+5\)
22) \(\dfrac{1}{5x^2}+\dfrac{1}{x^2-9x+36}=\dfrac{1}{x^2-4x+16}\)
23) \(\dfrac{\left(x+1\right)^2}{x^2+2x+2}-\dfrac{x^2+2x}{\left(x+1\right)^2}=\dfrac{1}{90}\)
24) \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
25)\(\dfrac{x-4}{x+1}+\dfrac{x-4}{x+1}+\dfrac{8}{3}=\dfrac{x-8}{x+2}+\dfrac{x+8}{x-2}\)
Thanks các cậu vì đã giúp mk
210. Giải phương trình:
\(\left|x+1\right|+3\left|x-1\right|=x+2+\left|x\right|+2\left|x-2\right|\)
Bài tập: Giải các phương trình sau:
a) \(\frac{1}{x+1}\)- \(\frac{5}{x-2}\)=\(\frac{15}{\left(x+1\right)\left(2-x\right)}\)
b)\(\frac{1}{2x-3}\)-\(\frac{3}{x\left(2x-3\right)}\)=\(\frac{5}{x}\)
c)\(\frac{6}{x-1}\)-\(\frac{4}{x-3}\)=\(\frac{8}{2x-6}\)
d)\(\frac{3}{\left(x-1\right)\left(x-2\right)}\)+\(\frac{2}{\left(x-3\right)\left(x-1\right)}\)=\(\frac{1}{\left(x-2\right)\left(x-3\right)}\)
